Question

. / A random sample of 8 cigarettes of a certain brand has an average nicotine content of 2.6 milligram and standard deviation of 0.9 milligrams. Construct a 99% confidence interval for the true average nicotine content of this brand of cigarettes, assuming the distribution of nicotine content to be normal.

Answer 1

To construct a 99% confidence interval for the true average nicotine content based on the given sample data, we will use the t-distribution because the sample size is small (n < 30) and the population standard deviation is unknown.

### Step 1: Gather the data
- Sample mean (\( \bar{x} \)): 2.6 mg
- Sample standard deviation (\( s \)): 0.9 mg
- Sample size (\( n \)): 8

### Step 2: Determine the t-value
To find the t-value for a 99% confidence interval, we need to know the degrees of freedom:
\[
\text{Degrees of freedom (df)} = n - 1 = 8 - 1 = 7
\]
Using a t-table or calculator for a 99% confidence level and 7 degrees of freedom, we find that the critical value \( t \) is approximately 3.499.

### Step 3: Calculate the Standard Error (SE)
The standard error of the mean is calculated as:
\[
SE = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{8}} \approx \frac{0.9}{2.8284} \approx 0.3182
\]

### Step 4: Calculate the Margin of Error (ME)
The margin of error is calculated using the formula:
\[
ME = t \times SE = 3.499 \times 0.3182 \approx 1.1131
\]

### Step 5: Construct the Confidence Interval
Now we can calculate the confidence interval:
\[
\text{Lower limit} = \bar{x} - ME = 2.6 - 1.1131 \approx 1.4869
\]
\[
\text{Upper limit} = \bar{x} + ME = 2.6 + 1.1131 \approx 3.7131
\]

### Step 6: Present the final result
Thus, the 99% confidence interval for the true average nicotine content of this brand of cigarettes is approximately:
\[
(1.49, 3.71) \text{ mg}
\]

So the final confidence interval is:
\[
(1.49, 3.71) \text{ mg}
\]